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IB Math – Standard Level – Calculus Practice Problems - Markscheme Alei - Desert Academy

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Diagrams for Calc Practice problems (they don’t print from Mac when text wrapping is on).

10

4

2

–2

–4

–6

–8

–10

–12

–1 1 2 3 4 5 6x

y

P(3, 2)

13x

y

0

17

f

g

p

x

y

16

12

8

4

0.5 1 1.5 18x

y

16

12

8

4

0.5 1 1.5

5

a

19 25

4

3

2

11 1

3

y = 1+1x–

0 1 2 3 4

27

A

C

B

y

x

O

34x

y

43π π



35

–1 0 1 2

A

B

C D

y

x

37

6

4

2

1 2 3–2 –1

D

C(a, b)

y

x

39

y

4

4

2

20

–2

–2–4 x

40 a

5

4

3

2

1

–1

–2

–3

–4

–5

–5 –4 –3 –2 –1 21 3 4 50 x

y

40b

5

4

3

2

1

–1

–2

–3

–4

–5

–5 –4 –3 –2 –1 21 3 4 50 x

y

A

43

R

1

2

–1

–1

112

12

–x

y

44

–5 –4 –3 –2 –1 1 2 3 4 5 x

y

S

( , 0)b( , 0)a

R

4

3

2

1

–1

–2

–3

–4

–5

–6

–7 50



SL Calculus Practice Problems - MarkScheme 1. y = sin (2x – 1)

x

y

d

d = 2 cos (2x – 1) (A1)(A1)

At

0,2

1, the gradient of the tangent = 2 cos 0 (A1)

= 2 (A1) (C4)

2. (a) (i) a = –3 (A1)

(ii) b = 5 (A1) 2

(b) (i) f ′(x) = –3x2

+ 4x + 15

(A2)

(ii) –3x2

+ 4x + 15 = 0

–(3x + 5)(x – 3) = 0 (M1)

x = –3

5 or x = 3 (A1)(A1)

OR

x = –3

5 or x = 3 (G3)

(iii) x = 3 ⇒ f (3) = –33

+ 2(32) + 15(3) (M1)

= –27 + 18 + 45 =36 (A1)

OR

f (3) = 36 (G2) 7

(c) (i) f ′(x) = 15 at x = 0 (M1)

Line through (0, 0) of gradient 15

⇒ y = 15x (A1)

OR

y = 15x (G2)

(ii) –x3

+ 2x2

+ 15x = 15x (M1)

⇒ –x3

+ 2x2

= 0

⇒ –x2

(x – 2) = 0

⇒ x = 2 (A1)

OR

x = 2 (G2) 4

(d) Area =115 (3 sf) (G2)

OR

Area = ∫

++−=++−

6

0

5

0

23423

215

32

4d)152(

xxxxxxx (M1)

= 12

1375 = 115 (3 sf) (A1) 2

[15]

3. (a) (i) v(0) = 50 – 50e0 = 0 (A1)

(ii) v(10) = 50 – 50e–2

= 43.2 (A1) 2

(b) (i) a = t

v

d

d = –50(–0.2e

–0.2t) (M1)

= 10e–0.2t

(A1)

(ii) a(0) = 10e0 = 10 (A1) 3

(c) (i) t → ∞ ⇒ v → 50 (A1)

(ii) t → ∞ ⇒ a → 0 (A1)

(iii) when a = 0, v is constant at 50 (R1) 3

(d) (i) y = ∫vdt (M1)



= 50t –2.0

e –0.2

−

t

+ k (A1)

= 50t + 250e–0.2t

+ k (AG)

(ii) 0 = 50(0) + 250e0

+ k = 250 + k (M1)

⇒ k = –250 (A1)

(iii) Solve 250 = 50t + 250e–0.2t

– 250 (M1)

⇒ 50t + 250e–0.2t

– 500 = 0

⇒ t + 5e–0.2t

– 10 = 0

⇒ t = 9.207 s (G2) 7

[15]

4. METHOD 1

0 x

y

Using gdc coordinates of maximum are

(0.667, 26.9) (G3)(G3)(C6)

METHOD 2

At maximum x

y

d

d = 3x

2 – 20x + 12 = 0 = (3x – 2) (x – 6) (M1)(A1)(M1)

=> x = 3

2 must be where maximum occurs (A1)

x = 3

2 => y =

3

3

2

– 10

2

3

2

+ 12

3

2 + 23 =

27

725 (= 26.9, 3 sf) (M1)(A1)

Maximum at

27

725,

3

2 (C4)(C2)

[6]

5. (a) t

s

d

d= 30 – at => s = 30t – a

2

2t + C (A1)(A1)(A1)

Note: Award (A1) for 30t, (A1) for a2

2t, (A1) for C.

t = 0 => s = 30(0) – a( )2

02

+ C = 0 + C => C = 0 (M1)

=> s = 30t – 2

1at

2 (A1) 5

(b) (i) vel = 30 – 5(0) = 30 m s–1

(A1)

(ii) Train will stop when 0 = 30 – 5t => t = 6 (M1)

Distance travelled = 30t – 2

1at

2

= 30(6) – 2

1(5) (6

2) (M1)

= 90m (A1)

90 < 200 => train stops before station. (R1)(AG) 5



(c) (i) 0 = 30 – at => t = a

30 (A1)

(ii) 30

a

30–

2

1(a)

230

a = 200 (M1)(M1)

Note: Award (M1) for substituting a

30, (Ml) for setting equal to 200.

=> aaa

450450–

900= = 200 (A1)

=> a = 4

9

200

450= = 2.25 m s

–2 (A1) 5

Note: Do not penalize lack of units in answers.

[15]

6. (i) At x = a, h (x) = a 5

1

h′ (x) = 5

1x 5

4–

=> h′ (a) =

5

4

5

1

a

= gradient of tangent (A1)

=> y – a 5

1

=

5

4

5

1

a

(x – a) =

5

4

5

1

a

x – 5

1a 5

1

(M1)

=> y =

5

4

5

1

a

x + 5

4a 5

1

(A1)

(ii) tangent intersects x-axis => y = 0

=>

5

4

5

1

a

x = –5

4a 5

1

(M1)

=> x = 5a 5

4

5

1

5

4– a = –4a (M1)(AG) 5

[5]

7. (a) (i) When t = 0, v = 50 + 50e0

(A1)

= 100 m s–1

(A1)

(ii) When t = 4, v = 50 + 50e–2

(A1)

= 56.8 m s–1

(A1) 4

(b) v = t

s

d

d

⇒ s = ∫ tv d

( ) tt∫ +4

0

0.5- de5050 (A1)(A1)(A1) 3

Note: Award (A1) for each limit in the correct position and (A1) for the function.

(c) Distance travelled in 4 seconds = ( ) tt∫ +4

0

0.5- de5050

= [50t – 100e–0.5t

]4

0 (A1)

= (200 – 100e–2

) – (0 – 100e0)

= 286 m (3 sf) (A1)

Note: Award first (A1) for [50t – 100e–0.5t

], ie

limits not required.



OR

Distance travelled in 4 seconds = 286 m (3 sf) (G2) 2

(d)

100

50

velocity

(t = 4)

121086420 t

v

time Notes: Award (A1) for the exponential part, (A1) for the straight line through (11, 0),

Award (A1) for indication of time on x-axis and velocity on

y-axis,

(A1) for scale on x-axis and y-axis.

Award (A1) for marking the point where t = 4.

5

(e) Constant rate = 7

8.56 (M1)

= 8.11 m s–2

(A1) 2

Note: Award (M1)(A0) for –8.11.

(f) distance = 2

1(7)(56.8) (M1)

= 199 m (A1) 2

Note: Do not award ft in parts (e) and (f) if candidate has not used a straight line for t = 4 to t

= 11 or if they continue the exponential beyond t = 4.

[18]

8. y

x

(A2)(A1)(A1)(A2) (C6)

Note: Award A2 for correct shape (approximately parabolic), A1 A1 for intercepts at 0 and 4, A2

for minimum between x = 1.5 and x = 2.5.

[6]



9. (a) (i) f ′(x) = –2e–2x

(A1)

(ii) f ′(x) is always negative (R1) 2

(b) (i) y = 1 + 2

1–2–

e×

(= 1 + e) (A1)

(ii) f ′ 2

1–2–

e22

1 ×−=

− (= –2e) (A1) 2

Note: In part (b) the answers do not need to be simplified.

(c) y – (1 + e) = –2e

+

2

1x (M1)

y = –2ex + 1 ( y = –5.44 x + 1) (A1)(A1) 3

(d) (i) (ii) (iii)

P

(A1)(A1)(A1)

Notes: Award (A1) for each correct answer. Do not allow (ft) on an incorrect answer to part (i).

The correct final diagram is shown below. Do not penalize if the horizontal asymptote is missing.

Axes do not need to be labelled.

(i)(ii)(iii)

–1 1 2x

y

8

6

4

2

1

P

12

–

(iv) Area = ∫−− +−−+

0

2

12 d)]1e2()e1[( xxx

(or equivalent) (M1)(M1)

Notes: Award (M1) for the limits, (M1) for the function.

Accept difference of integrals as well as integral of difference. Area below line may be calculated

geometrically.

Area = ∫−− +

0

2

12 d)e2e[( xxx

=

0

2

1

22 ee2

1

−

−

+− xx

(A1)

= 0.1795 …= 0.180 (3 sf) (A1)

OR

Area = 0.180 (G2) 7

[14]



10. (a) x = 1 (A1)1

(b) (i) f (–1000) = 2.01 (A1)

(ii) y = 2 (A1) 2

(c) f ′(x) = 4

22

)1(

)20132)(1(2)134()1(

−+−−−−−

x

xxxxx (A1)(A1)

= 3

22

)1(

)40264()13174(

−+−−+−

x

xxxx (A1)

= 3)1(

279

−−

x

x (AG) 3

Notes: Award (M1) for the correct use of the quotient rule, the first (A1) for the placement of the

correct expressions into the quotient rule.

Award the second (A1) for doing sufficient simplification to make the given answer reasonably

obvious.

(d) f ′(3) = 0 ⇒ stationary (or turning) point (R1)

f ″(3) = 16

18 > 0 ⇒ minimum (R1) 2

(e) Point of inflexion ⇒ f ″(x) = 0 ⇒ x = 4 (A1)

x = 4 ⇒ y = 0 ⇒ Point of inflexion = (4, 0) (A1)

OR

Point of inflexion = (4, 0) (G2) 2

[10]

11. (a) d = ∫ −+4

0

– )e554( tt dt (M1)(A1)(A1)(C3)

Note: Award (M1) for ∫, (A1) for both limits, (A1) for 4t + 5 – 5e–t

(b) d = 40

2 ]e552[ -ttt ++ (A1)(A1)

Note: Award (A1) for 2t2

+ 5t, (A1) for 5e–t

.

= (32 + 20 + 5e–4

) – (5)

= 47 + 5e–4

(47.1, 3sf ) (A1) (C3)

[6]

12. (a) Velocity is d

d

s

t. (M1)

d10

d

st

t= − (A1)

10 (m s–1

) (A1) (C3)

(b) The velocity is zero when d

0d

s

t= (M1)

10 0t− =

10t = (secs) (A1)(C2)

(c) s = 50 (metres) (A1) (C1)

Note: Do not penalize absence of units.

[6]

13. (a) 3h = (A1)

2k = (A1) 2

(b) ( )f x2( 3) 2x= − − +

2 6 9 2x x= − + − + (must be a correct expression) (A1)

2 6 7x x= − + − (AG)1

(c) ( ) 2 6f x x′ = − + (A2) 2

(d) (i) tangent gradient 2= − (A1)



gradient of L 1

2= (A1) (N2) 2

(ii) EITHER

equation of L is 1

2y x c= + (M1)

1c = − . (A1)

11

2y x= −

OR

11 ( 4)

2y x− = − (A2) (N2) 2

(iii) EITHER

2 16 7 1

2x x x− + − = − (M1)

22 11 12 0x x− + = (may be implied) (A1)

(2 3)( 4) 0x x− − = (may be implied) (A1)

1.5x = (A1) (N3) 4

OR

2 16 7 1

2x x x− + − = − (or a sketch) (M1)

1.5x = (A3) (N3) 8

[13]

14. (a) 4x = (A1)

g″ changes sign at 4x = or concavity changes (R1) 2

(b) 2x = (A1)

EITHER

g′ goes from negative to positive (R1)

OR

g′ (2) = 0 and g″ (2) is positive (R1) 2

(c)

1 2 3 4 5 6 7 8

P

M

(A2)(A1)(A1) 4

Note: Award (A2) for a suitable cubic curve through (4, 0), (A1) for M at x = 2, (A1) for P at (4,

0).

[8]



15. (a) tva

dd= (M1)

= –10 A1 3

(b) s = ∫vdt (M1)

= 50t – 5t2

+ c A1

40 = 50(0) – 5(0) + c ⇒ c = 40 A1

s = 50t – 5t2

+ 40 A1 3

Note: Award (M1) and the first (A1) in part (b) if c is missing, but do not award the final 2

marks.

[6]

16. (a) (i) f ′(x) = –x + 2 A1

(ii) f ′(0) = 2 A12

(b) Gradient of tangent at y-intercept = f ′(0) = 2

⇒ gradient of normal = 21

(= –0.5) A1

Finding y-intercept is 2.5 A1

Therefore, equation of the normal is

y – 2.5 = ~(x – 0) (y – 2.5 = –0.5x) M1

(y = –0.5x + 2.5 (AG) 3

(c) (i) EITHER

solving –0.5x2

+ 2x + 2.5 = –0.5x + 2.5 (M1)A1

⇒ x = 0 or x = 5 A1 2

OR

f x( )

y

x

g x( )

M1

Curves intersect at x = 0, x = 5 (A1)

So solutions to f (x) = g (x) are x = 0, x = 5 A12

OR

⇒ 0.5x2

– 2.5x = 0 (A1)

⇒ – 0.5x(x – 5) = 0 M1

⇒ x = 0 or x = 5 A12

(ii) Curve and normal intersect when x = 0 or x = 5 (M2)

Other point is when x = 5

⇒ y = –0.5(5) + 2.5 = 0 (so other point (5, 0) A1 2

(d) (i) Area = ∫ ∫

××−++−−

5

0

5

0

2 5.2521d)5.225.0(or d))()(( xxxxxgxf

A1A1A1 3

Note: Award (A1) for the integral, (A1) for both correct

limits on the integral, and (A1) for the difference.

(ii) Area = Area under curve – area under line (A = A1 – A2) (M1)

(A1) = 425,

350

2 =A

Area = 12125

425

350 =− (or 10.4 (3sf) A12

[16]



17. (a) (i) p = (10x + 2) – (1 + e2x

) A22

Note: Award (A1) for (l + e2x

) – (10x + 2).

(ii) x

p

d

d= 10 – 2e

2x A1A1

x

p

d

d = 0 (10 – 2e

2x = 0) M1

x = 2

5n1 (= 0.805) A14

(b) (i) METHOD 1

x = 1 + e2x

M1

1n(x – 1) = 2y A1

f –1

(x) =

−

=−

2

)1(n1 Allow

2

)1(n1 xy

x A13

METHOD 2

y – 1 = e2x

A1

2

)1ln( −y = x M1

f –1

(x) =

−

=−

2

)1(n1 Allow

2

)1(n1 xy

x A13

(ii) a =

=

− 22n121

2

)15(n1 M1

= 21 × 21n2 A1

= 1n 2 AG2

(c) Using V = xyb

adπ 2∫ (M1)

Volume =

++ ∫∫

805.0

0

222ln

0

22 d)e1(πord)e1(π xx xx A2 3

[14]

18. (a) (i) 2p = − 4q = (or 4, 2p q= = − ) (A1)(A1)(N1)(N1)

(ii) ( 2)( 4)y a x x= + −

8 (6 2)(6 4)a= + − (M1)

8 16a=

1

2a = (A1)(N1)

(iii) 1

( 2)( 4)2

y x x= + −

21( 2 8)

2y x x= − −

214

2y x x= − − (A1) (N1) 5

(b) (i) d

1d

yx

x= − (A1) (N1)

(ii) 1 7x − = (M1)

( )8, 20 P is (8, 20)x y= = (A1)(A1) (N2) 4

(c) (i) when x = 4, gradient of tangent is 4 – 1 = 3 (may be implied) (A1)



gradient of normal is 1

3− (A1)

1 1 40 ( 4)

3 3 3y x y x

− = − − = − +

(A1)(N3)

(ii) 21 1 4

42 3 3

x x x− − = − + (or sketch/graph) (M1)

21 2 160

2 3 3x x− − =

23 4 32 0x x− − = (may be implied) (A1)

(3 8)( 4) 0x x+ − =

8or 4

3x x= − =

8( 2.67)

3x = − − (A1) (N2) 6

[15]

19. METHOD 1 l + 2w = 60 (M1)

l = 60 − 2w (A1)

A = w(60 − 2w) (= 60w − 2w2

) (A1)

w

A

d

d = 60 − 4w (A1)

Using w

A

d

d = 0 (60 − 4w = 0) (M1)

w = 15 (A1) (C6)

METHOD 2 w + 2l = 60 (A1)

w = 60 − 2l (A1)

A = l(60 − 2l) (= 60l − 2l2

) (A1)

l

A

d

d = 60 − 4l (A1)

Using l

A

d

d = 0 (60 − 4l = 0) (M1)

l = 15

w = 30 (A1) (C6)

[6]

20. (a) s = 25t − ct +3

3

4 (M1)(A1)(A1)

Note: Award no further marks if “c” is

missing.

Substituting s = 10 and t = 3 (M1)

10 = 25 × 3 − c+3(3)3

4

10 = 75 − 36 + c

c = − 29 (A1)

s = 25t − 293

4 3 −t (A1) (N3)

(b) METHOD 1

s is a maximum when v = 0d

d=

t

s (may be implied) (M1)



25 − 4t2

= 0 (A1)

t2

= 4

25

t = 2

5 (A1) (N2)

METHOD 2

Using maximum of s (3

212 , may be implied) (M1)

25t − 3

21229

3

4 3 =−t (A1)

t = 2.5 (A1) (N2)

(c) 25t − 0293

4 3 >−t (accept equation) (M1)

m = 1.27, n = 3.55 (A1)(A1) (N3)

[12]

21. f ′(x) = cos x ⇒ f (x) = sin x + C (M1)

f

2

π = –2 ⇒ –2= sin

2

π + C (M1)

C = –3 (A1)

f (x) = sin x – 3 (A1) (C4)

[4]

22. (a) y = π sin x – x

(–2.3, 0)

(–1.25, –1.73)

(1.25, 1.73)

(2.3, 0)

–3

–3

–2

–2

–1

–1

1

1

2

2

3

3

y

x

(A5)5

Notes: Award (A1) for appropriate scales marked on the axes.

Award (A1) for the x-intercepts at (±2.3, 0).

Award (A1) for the maximum and minimum points at (±1.25, ±1.73).

Award (A1) for the end points at (±3, ±2.55).

Award (A1) for a smooth curve.

Allow some flexibility, especially in the middle three marks here.

(b) x = 2.31 (A1) 1

(c) ∫ +−−=− Cx

xxxx2

cosπd)sinπ(2

(A1)(A1)

Note: Do not penalize for the absence of C.

Required area = ∫ −1

0d)sinπ( xxx (M1)

= 0.944 (G1)

OR area = 0.944 (G2) 4

[10]



23. f ′(x) = –2x + 3

f (x) = 2

2 2x− + 3x + c (M1)

Notes: Award (M1) for an attempt to integrate. Do not penalize the omission of c here.

1 = –1 + 3 + c (A1)

c = –1 (A1)

f (x) = –x2

+ 3x – 1 (A1) (C4)

[4]

24. (a) f ′(x) = 3(2x + 5)2 × 2 (M1)(A1)

Note: Award (M1) for an attempt to use the chain rule.

= 6(2x + 5)2 (C2)

(b) ∫ ×+

=24

)52(d)(

4xxxf + c (A2) (C2)

Note: Award (A1) for (2x + 5)4

and (A1) for /8.

[4]

25.

2

1

Area = ∫∫ −=

2

3

11

2

3

11

d)1(

1d y

yyx (M1)(A1)

= [ ]23

11

)1(ln −y

= ln 1 – ln3

1 (A1)

= ln 3 (A1) (C4)

OR

Area from x = 1 to x = 3, A = ∫ +=

+

3

1

31]ln[d

11 xxx

x

= (3 + ln 3) – (1 + ln 1) (M1)

= 2 + ln 3 (A1)

Area rectangle 1

= 2 × 13

1 = 2

3

2, area rectangle

2 = 1 ×

3

2

3

2=

Shaded area = 2 + ln 3 – 23

2

3

2+ (M1)

= ln 3 (A1) (C4)

OR

Area from x = 1 to x = 3, A = xx

d1

13

1∫

+ (M1)

A = 3.0986 … (G0)



Area rectangle 1

= 2 × 13

1 = 2

3

2, area rectangle

2 = 1 ×

3

2

3

2=

Shaded area = 3.0986 – 2 3

2

3

2+ (M1)

= 1.10 (3 sf) (A1) (C4)

Notes: An exact value is required. If candidates have obtained the answer 1.10, and shown their

working, award marks as above. However, if they do not show their working, award (G2) for the

correct answer of 1.10.

Award no marks for the giving of 3.10 as the final answer.

[4]

26. (a)(i) & (c)(i)

1

0

–1

–2

1 2

(1.1, 0.55)

R

(1.51, 0)

(2, –1.66)

y

x

(A3)

Notes: The sketch does not need to be on graph paper. It should have the correct shape, and the

points (0, 0), (1.1, 0.55), (1.57, 0) and (2, –1.66) should be indicated in some way.

Award (A1) for the correct shape.

Award (A2) for 3 or 4 correctly indicated points, (A1) for 1 or 2 points.

(ii) Approximate positions are

positive x-intercept (1.57, 0) (A1)

maximum point (1.1, 0.55) (A1)

end points (0, 0) and (2, –1.66) (A1)(A1) 7

(b) x2

cos x = 0 x ≠ 0 � cos x = 0 (M1)

⇒ x = 2

π (A1) 2

Note: Award (A2) if answer correct.

(c) (i) see graph (A1)

(ii) ∫π2

0

2x cos x dx (A2) 3

Note: Award (A1) for limits, (A1) for rest of integral correct (do not penalize missing dx).

(d) Integral = 0.467 (G3)

OR

Integral = [ ] 2/π

02 sin2cos2sin xxxxx −+ (M1)

=

−

+ )1(2)0(

2

π2)1(

4

π 2

– [0 + 0 – 0] (M1)

= 2

π – 2 (exact) or 0.467 (3 sf) (A1) 3

[15]



27. (a) From graph, period = 2π (A1)1

(b) Range = {y |–0.4 < y < 0.4} (A1) 1

(c) (i) f ′(x) = xd

d{cos x (sin x)

2}

= cos x (2 sin x cos x) – sin x (sin x)2

or –3 sin3

x + 2 sin x (M1)(A1)(A1)

Note: Award (M1) for using the product rule and (A1) for each part.

(ii) f ′(x) = 0 (M1)

⇒ sin x{2 cos x – sin2

x} = 0 or sin x{3 cos x – 1} = 0 (A1)

⇒ 3 cos2

x – 1 = 0

⇒ cos x = ±

3

1 (A1)

At A, f (x) > 0, hence cos x =

3

1 (R1)(AG)

(iii) f (x) =

2

3

1–1

3

1 (M1)

= 39

2

3

1

3

2=× (A1) 9

(d) x = 2

π (A1) 1

(e) (i) cxxxx +=∫ 32 sin3

1d))(sin(cos (M1)(A1)

(ii) Area = ∫

−

=

2/π

0

3

3

2 )0(sin2

πsin

3

1d))(sin(cos xxx (M1)

= 3

1 (A1) 4

(f) At C f ″(x) = 0 (M1)

⇔ 9 cos3

x – 7 cos x = 0

⇔ cos x(9 cos2 x – 7) = 0 (M1)

⇒ x = 2

π (reject) or x = arccos

3

7 = 0.491 (3 sf) (A1)(A1) 4

[20]

28. f ′(x) = 1 – x2

f (x) = Cx

xxx +−=−∫ 3d)1(

32

(A1)

f (3) = 0 ⇒ 3 – 9 + C = 0 (M1)

⇒ c = 6 (A1)

f (x) = x – 3

3x + 6 (A1)

[4]



29. (a)

4

3

2

1

1 2 3 4 5

–1

integerson axis

0.5< <13.5< <4

xy

3.2< <3.6–0.2< <0

xy

{

{

{

(A1)

(A1) (A1)

(A1)

(A1)

LEFT

INTERCEPT

RIGHT

INTERCEPT3< <3.5x 3.5< <4x

MAXIMUM

POINT

MINIMUMPOINT

y

x

5

(b) π is a solution if and only if π + π cos π = 0. (M1)

Now π + π cos π = π + π(–1) (A1)

= 0 (A1) 3

(c) By using appropriate calculator functions x = 3.696 722 9... (M1)

⇒ x = 3.69672 (6sf) (A1) 2

(d) See graph: (A1)

∫ +π

0d)cosπ( xxx (A1) 2

(e) EITHER ∫ +π

0d)cosπ( xxx = 7.86960 (6 sf) (A3) 3

Note: This answer assumes appropriate use of a calculator eg

‘fnInt’:

+==

xxYwith

XYfnInt

cosπ

869604401.7)π,0,,(

1

1

OR π0

π

0]cossinπ[d)cosπ( xxxxxxx ++=+∫

= π(π – 0) + (π sin π – 0 × sin 0) + (cos π – cos 0) (A1)

= π2 + 0 + –2 = 7.86960 (6 sf) (A1) 3

[15]

30. Note: Do not penalize for the omission of C.

(a) Cxxx ++−=+∫ )73(cos3

1d)73(sin (A1)(A1) (C2)

Note: Award (A1) for 3

1, (A1) for –cos (3x + 7).

(b) ∫ −=−

4

1de 4 xx

e–4x

+ C (A1)(A1) (C2)

Note: Award (A1) for –4

1, (A1) for e

–4x.

[4]

31. f (x) = ∫

+xx

xd sin5.0–

1

1 (M1)

= ln x + 1 + 0.5 cos x + c (A1)(A1)(A1)

2 = ln 1 + 0.5 + c (M1)

c = 1.5 (A1)

f (x) = ln x + 1 + 0.5 cos x + 1.5 (C6)

[6]



32. (a)

1

0

1

0

-e1

d-e∫

−= kxkx

kx (A1)

= –k

1(e

–k – e

0) (A1)

= –k

1 (e

–k – 1) (A1)

= –k

1 (1 – e

–k) (AG)3

(b) k = 0.5

(i)

–1 0 1 2 3

1

(0,1)

y

x

(A2)

Note: Award (A1) for shape, and (A1) for the point (0,1).

(ii) Shading (see graph) (A1)

(iii) Area = ∫1

0d-e xkx

for k = 0.5 (M1)

= 5.0

1(1 – e

0.5)

= 0.787 (3 sf) (A1)

OR

Area = 0.787 (3 sf) (G2) 5

(c) (i) x

y

d

d = –ke

–kx (A1)

(ii) x = 1 y = 0.8 ⇒ 0.8 = e –k

(A1)

ln 0.8 = –k

k = 0.223 (A1)

(iii) At x = 1 x

y

d

d = –0.223e

–0.223 (M1)

= –0.179 (accept –0.178) (A1)

OR

x

y

d

d = –0.178 or – 0.179 (G2) 5

[13]

33. f (x) = 2

3

x (M1)

(a) f ′(x) = 1–

2

3

2

3x = 2

1

2

3x (or x

2

3) (M1)(A1) (C3)

(b) ∫ xx d2

3

= cx ++

+12

3

12

3

1 (M1)



= cx +2

5

5

2 (or

5

5

2x + c) (A1)(A1) (C3)

Notes: Do not penalize the absence of c.

Award (A1) for 2

5 and (A1) for 2

5

x .

[6]

34. Area = ∫b

axx dsin (M1)

a = 0, b = 4

π3 (A1)

Area = ∫π4

3

0dsin xx = [–cos x] 4

π3

0 (A1)

=

4

π3cos– – (– cos 0) (A1)

= –

2

2– – (–1) (A1)

= 1 + 2

2 (A1)(C6)

Note: Award (G3) for a gdc answer of 1.71 or 1.707.

[6]

35. (a) At A, x = 0 => y = sin (e0

) = sin (1) (M1)

=> coordinates of A = (0,0.841) (A1)

OR

A(0, 0.841) (G2)2

(b) sin (ex) = 0 => e

x = π (M1)

=> x = ln π (or k = π) (A1)

OR

x = ln π (or k = π) (A2) 2

(c) (i) Maximum value of sin function = 1 (A1)

(ii) x

y

d

d = e

x cos (e

x) (A1)(A1)

Note: Award (A1) for cos (ex) and (A1) for e

x.

(iii) x

y

d

d= 0 at a maximum (R1)

ex

cos (ex) = 0

=> ex

= 0 (impossible) or cos (ex) = 0 (M1)

=> ex =

2

π => x = ln

2

π (A1)(AG) 6

(d) (i) Area = ∫πln

0d)(esin xx

(A1)(A1)(A1)

Note: Award (A1) for 0, (A1) for ln π, (A1) for sin (ex).

(ii) Integral = 0.90585 = 0.906 (3 sf) (G2) 5

(e)



y x = 3

p

(M1)

At P, x = 0.87656 = 0.877 (3 sf) (G2)3

[18]

36. (a) 2

1 × 10 = 5 (M1)(A1)(C2)

(b) ( )∫ ∫+3

1

3

1d4d xxxg (M1)

[ ]∫3

1

3

14d4 xx (A1)

= 4 × 2 = 8 (A1)

( )( ) xxg d43

1∫ + = 10 + 8 = 18 (A1) (C4)

[6]

37. (a) (i) cos 2

1

4

π– =

, sin 2

1–

4

π– =

(A1)

therefore cos

+

4

π–sin

4

π– = 0 (AG)

(ii) cos x + sin x = 0 ⇒ 1 + tan x = 0

⇒ tan x = –l (M1)

x = 4

π3 (A1)

Note: Award (A0) for 2.36.

OR

x = 4

π3 (G2) 3

(b) y = ex(cos x + sin x)

x

y

d

d = e

x(cos x + sin x) + e

x(–sin x + cos x) (M1)(A1)(A1) 3

= 2ex cos x

(c) x

y

d

d = 0 for a turning point ⇒ 2e

x cos x = 0 (M1)

⇒ cos x = 0 (A1)

⇒ x = 2

π ⇒ a =

2

π (A1)

y = e 2

π

(cos 2

π + sin

2

π) = e 2

π

b = e 2

π

(A1) 4

Note: Award (M1)(A1)(A0)(A0) for a = 1.57, b = 4.81.



(d) At D, 2

2

d

d

x

y= 0 (M1)

2ex cos x – 2e

xsin x = 0 (A1)

2ex (cos x – sin x) = 0

⇒ cos x – sin x = 0 (A1)

⇒ x = 4

π (A1)

⇒ y = e 4

π

(cos 4

π + sin

4

π) (A1)

= 2 e 4

π

(AG) 5

(e) Required area = ∫ 4

3

0e

x(cos x + sin x)dx (M1)

= 7.46 sq units (G1)

OR

Αrea = 7.46 sq units (G2) 2

Note: Award (M1)(G0) for the answer 9.81 obtained if the calculator is in degree mode.

[17]

38. y = ∫ x

y

d

ddx (M1)

= 2

2

4

24 xx+ – x + c (A1)(A1)

Note: Award (A1) for first 3 terms, (A1) for “+ c”.

13 = 4

16 + 4 – 2 + c (M1)

c = 7 (A1)

y = 4

4x + x

2 – x – 7 (A1) (C6)

[6]

39. (a) ∫ ++ xx 2))d (sin 3 (1 = x – 3 cos (x + 2) + c (A1)(A1)(A1)(C3)

Notes: Award A1 for x, A1 for –cos (x + 2) A1 for coefficient 3,

ie A1 A1 for the second term, which may be written as

+3(–cos (x + 2))

Do not penalize the omission of c.

(b) 1 + 3 sin (x + 2) = 0 (M1)

sin (x + 2) = –3

1

x + 2 = –0.3398, π + 0.3398, ... (A1)

x = –2.3398, 1.4814, ...

Required value of x = 1.48 (A1) (C3)

[6]

40. (a)

x

y

(A1)(A1)2



Note: Award (A1) for a second branch in approximately the correct position, and (A1) for the

second branch having positive x and y intercepts. Asymptotes need not be drawn.

(b) (i) x-intercept =

=

2

1,0,

2

1Accept

2

1x (A1)

y-intercept = 1 (Accept (0, 1), y = 1) (A1)

(ii) horizontal asymptote y = 2 (A1)

vertical asymptote x = 1 (A1)4

(c) (i) f ′(x) = 0 – (x – 1)–2

−−

=2)1(

1

x (A2)

(ii) no maximum / minimum points.

since 2)1(

1

−−

x ≠ 0 (R1)3

(d) (i) 2x + ln (x – 1) + c (accept lnx – 1) (A1)(A1)(A1)

(ii) A = [ ]∫ ∫

−+

−+

4

2

4

2

4

2)1(ln2,d

1

12Acceptd)( xxx

xxxf (M1)(A1)

Notes: Award (A1) for both correct limits.

Award (M0)(A0) for an incorrect function.

(iii) A = [ ]42)1(ln2 −+ xx

= (8 + ln 3) – (4 + ln 1) (M1)

= 4 + ln 3(= 5.10, to 3 sf) (A1) (N2)7

[16]

41. 21

( ) e ln (1 )2

xf x x c−= − − − + (M1)(A1)(A1)

Substituting 2(0)1

4 e ln(1 0)2

c−= − − − + 1

or 4 ln12

c = − − +

(M1)

4.5c = (A1)

21( ) e ln(1 ) 4.5

2

xf x x−= − − − + (A1)(C2)(C2)(C2)

[6]

42. (a) (i) 16 (A2)(C2)

(ii) 3 3

0 0( )d 2df x x x+∫ ∫ (or appropriate sketch) (M1)

14= (A1)(C2)

(b) ( 2)d 8d

cf x x− =∫

2, 5c d= = (A2)(C2)

[6]

43. (a) (i) ( )1 accept (1 , 0)a = − π − π (A1)

(ii) ( )1 accept (1 , 0)b = + π + π (A1) 2

(b) (i) 1 2

2.14 1( )d ( )dh x x h x x

−−∫ ∫ (M1)(A1)(A1)

OR

1 2

2.14 1( )d ( )dh x x h x x

−+∫ ∫ (M1)(A1)(A1)

OR 1 1

2.14 2( )d ( )dh x x h x x

−+∫ ∫ (M1)(A1)(A1)

(ii) 5.141... ( 0.1585...)− −

= 5.30 (A2) 5



(c) (i) y = 0.973 (A1)

(ii) 0.240 0.973k− < < (A3) 4

[11]

44. (a) 0y = (A1)1

(b) 2 2

2( )

(1 )

xf x

x

−′ =+

(A1)(A1)(A1) 3

(c)

2

2 3

6 20

(1 )

x

x

−=

+ (or sketch of ( )f x′ showing the maximum) (M1)

26 2 0x − = (A1)

1

3x = ± (A1)

1( 0.577)

3x

−= = − (A1) (N4) 4

(d) 0.5 0.5 0

2 2 20.5 0 0.5

1 1 1d = 2 d = 2 d

1 1 1x x x

x x x− −

+ + + ∫ ∫ ∫ (A1)(A1) 2

[10]

45. (a) Using the chain rule (M1)

f ′(x) =(2 cos (5x–3)) 5 (= 10 cos (5x – 3)) A1

f ″(x) = –(10 sin (5x–3)) 5

= –50 sin (5x – 3) A1A14

Note: Award (A1) for sin (5x – 3), (A1) for –50.

(b) ∫f (x)dx = 52

cos (5x – 3) + c A1A1 2

Note: Award (A1) for cos (5x–3), (A1) for 52− .

[6]

46. (a) ( )4 4( ) 5(3 4) 3 15(3 4)f x x x′ = + × = + (A1)(A1)(A1)(C3)

(b) 5(3 4) dx x+∫ =

661 1 (3 4)

(3 4)3 6 18

xx c c

+× + + = +

(A1)(A1)(A1) (C3)

[6]

47. Attempting to integrate. (M1)

3 5y x x c= − + (A1)(A1)(A1)

substitute (2, 6) to find c ( )36 2 5(2) c= − + (M1)

8c = (A1)

3 5 8y x x= − + (Accept

3 5 8x x− + ) (C6)

[6]

48. (a) ( ) ( )d( ) ( ) ( ) ( ) (4) (4)

df x g x f x g x f g

x′ ′′ ′+ = + = + (M1)

7 4= +

11= (A1)(C2)

(b) ( ) [ ] [ ]3 3 3

1 11( ) 6 d ( ) 6g x x g x x′ + = +∫ (A1)(A1)

( ) ( ) ( )(3) (1) 18 6 (2 1) 12g g= − + − = − + (A1)

13= (A1)(C4)

[6]

49. Using ∫ = xx

ln 1

(may be implied) (M1)



k

k

xxx

3

3

2)]([lnd2

1−=

−∫ (A1)

= ln (k − 2) − ln1 (A1)(A1)

ln (k − 2) − ln1 = ln 7

k − 2 = 7 (A1)

k = 9 (A1) (C6)

[6]

50. Note: There are many approaches possible.

However, there must be some evidence

of their method.

Area = ∫k

xx0

d2sin (must be seen somewhere) (A1)

Using area = 0.85 (must be seen somewhere) (M1)

EITHER

Integrating

k

x0

2cos2

1

−

+

−= 0cos

2

12kcos

2

1 (A1)

Simplifying 5.02cos2

1+

−k (A1)

Equation 5.02cos2

1+

−k = 0.85 (cos 2k = − 0.7)

OR Evidence of using trial and error on a GDC (M1)(A1)

Eg ∫π2

0d2sin xx = 0.5 ,

2

π too small etc

OR

Using GDC and solver, starting with ∫k

xx0

d2sin − 0.85 = 0 (M1)(A1)

THEN k = 1.17 (A2) (N3)

[6]